Chọn phương án B.

Đặt $A=\displaystyle\int\limits_1^3\dfrac{x+3}{x^2+3x+2}\mathrm{\,d}x$ ta có $$A=\ln2^a3^b5^c\Leftrightarrow\mathrm{e}^A=2^a3^b5^c.$$

Vì $a,\,b,\,c\in\mathbb{Z}$ nên $2^a3^b5^c\in\mathbb{Q}$.

1. Tính giá trị $A=\displaystyle\int\limits_1^3\dfrac{x+3}{x^2+3x+2}\mathrm{\,d}x$.
   
   
2. Khi đó $2^a3^b5^c$ bằng
   
   

Vì $\dfrac{12}{5}=4\cdot3\cdot\dfrac{1}{5}=2^2\cdot3^1\cdot5^{-1}$ nên $a=2$, $b=1$ và $c=-1$.

Khi đó $a+b+c=2$.

Chọn phương án B.

\(\begin{aligned}
\displaystyle\int\limits_1^3\dfrac{x+3}{x^2+3x+2}\mathrm{\,d}x&=\displaystyle\int\limits_1^3\dfrac{x+3}{(x+1)(x+2)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_1^3\left(\dfrac{2}{x+1}-\dfrac{1}{x+2}\right)\mathrm{\,d}x\\
&=\left(2\ln|x+1|-\ln|x+2|\right)\bigg|_1^3\\
&=2\ln2+\ln3-\ln5.\end{aligned}\)

Theo đó \(a=2,\,b=1,\,c=-1\).
Suy ra \(a+b+c=2\).